GHCI可以用來解釋強制嗎？

Question

如何讓GHCI解釋爲什麼以下幾點：

*Lib> sum Nothing 
0 

即使編譯？不知何故，有沒有Monoid？這不是簽名！

*Lib> :i Foldable 
class Foldable (t :: * -> *) where 
    ... 
    maximum :: Ord a => t a -> a 
    minimum :: Ord a => t a -> a 
    sum :: Num a => t a -> a 
    product :: Num a => t a -> a 
     -- Defined in ‘Data.Foldable’ 
instance Foldable [] -- Defined in ‘Data.Foldable’ 
instance Foldable Maybe -- Defined in ‘Data.Foldable’ 
instance Foldable (Either a) -- Defined in ‘Data.Foldable’ 
instance Foldable ((,) a) -- Defined in ‘Data.Foldable’ 
*Lib> :i Num 
class Num a where 
    (+) :: a -> a -> a 
    (-) :: a -> a -> a 
    (*) :: a -> a -> a 
    negate :: a -> a 
    abs :: a -> a 
    signum :: a -> a 
    fromInteger :: Integer -> a 
     -- Defined in ‘GHC.Num’ 
instance Num Word -- Defined in ‘GHC.Num’ 
instance Num Integer -- Defined in ‘GHC.Num’ 
instance Num Int -- Defined in ‘GHC.Num’ 
instance Num Float -- Defined in ‘GHC.Float’ 
instance Num Double -- Defined in ‘GHC.Float’ 
*Lib> sum Nothing 
0 

Answer 1

您可以使用類型孔：

> sum (Nothing :: _) 
<interactive>:4:17: 
    Found hole `_' with type: Maybe a 
    Where: `a' is a rigid type variable bound by 
       the inferred type of it :: Num a => a at <interactive>:4:1 
    To use the inferred type, enable PartialTypeSignatures 
    Relevant bindings include it :: a (bound at <interactive>:4:1) 
    In an expression type signature: _ 
    In the first argument of `sum', namely `(Nothing :: _)' 
    In the expression: sum (Nothing :: _) 

這表示，a是推斷出的類型的it :: Num a => a約束剛性類型變量，因爲Maybe是Foldable一個實例（如你已經發布在:i Foldable的輸出中，雖然你也可以在輸出中看到:i Maybe）編譯推定爲Nothing :: Num a => Maybe a，因爲sum的地方就是Num約束它。

那麼，爲什麼它編譯的原因是sum接受包含Num a => a值的Foldable和Maybe是Foldable，Nothing本身具有類型Maybe a和sum地方約束a必須實現Num。在GHCi中，默認爲Integer，因此您會看到0的輸出。